Prepared by Inam Jazbi – Learn Chemistry
- 🎯 Targeted Preparation: Covers key sections likely to appear in exams.
- 🔢 Important Numericals: Step-by-step solutions for scoring high marks.
- 🧪 Concept Clarity: Short-answer questions explained in simple language.
- ⏰ Smart Time Management: Suggested timings for each part to maximize efficiency.
- ✅ Exam-Oriented Approach: Structured exactly according to the paper pattern.
📝 Instructions
📚 Attempt 14 questions in all ---- 9 from Section B, 4 from Section C and the compulsory question No. 1 (Section A) of MCQs.
🔢 Write down proper numbering according to question paper. Do not copy questions.
✍️ Section B
🔹 Attempt 9 Short Questions ➡️ 4 marks each 🎯 Total = 36 marks
⏰ Suggested Time for Solution
🕒 Total: 90 mins
✍️ Each part: 9 mins
Q2 (i)
Define any 4 of the following:
Stoichiometry, Molar volume, Mole, Percentage yield, Exponential notation, Significant figure, Dipole moment, Tyndal effect, Anisotropy, Cleavage plane, Symmetry, Crystal growth, Transition temperature, Chemical equilibrium, Gay‑Lussac Law, Bond order, Debye, Unit cell, Bond energy, Colloid, Dispersion forces, Activation energy and Lattice energy, Rate expression, Rate constant, Critical temperature, Allotropy, Limiting reactant, Molar heat of vaporization, Dispersion force, Viscosity, Surface tension, Molar heat of fusion.
OR
Write down 2 differences between any TWO of the following:
➡️ Solution, colloids and suspension
➡️ Crystalline & amorphous solid
➡️ σ‑bond and π‑bond
➡️ Isomorphism and polymorphism
➡️ Ideal and non‑ideal solution
➡️ Polar and non‑polar bond
➡️ Continuous and line spectrum
➡️ Molecularity & reaction order
➡️ Lyman and Balmer series
➡️ VBT and MOT
➡️ Hydrophobic and Hydrophilic molecules
➡️ Positive and negative catalyst
➡️ BMO and AMO
Q2 (ii)
State Hund’s rule of multiplicity, Aufbau principle and Pauli exclusion principle. Write the values of four quantum numbers for the valence electrons of He and Mg (Z = 12). Also write down the electronic configuration for ground states of each of the following:
Zn, S²⁻, Cr (24), Cu (29), Fe³⁺ (26), Br⁻ (35), Mo (42), Ag (Z = 47), Pd (Z = 46), Ca²⁺ (Z = 20), Cl⁻, Sr²⁺ (38)
OR
What is meant by dipole moment? Give its mathematical formulae and different units. On what factors does it depend? Which of the following molecules have dipole moment? In each case, give a reason for your answer: CO₂, CHCl₃, SCl₂, H₂O, CCl₄
Q2 (iii)
What are quantum numbers and orbitals? Give a brief account of 4 quantum numbers. Write all possible values of l, m and s for n = 2 and n = 3. Draw the shape of different orbitals with l = 2. Arrange the following orbitals according to Wiswesser rule: 4f, 3d, 4s, 6p, 7s, 5d.
OR
What is bond energy? Give applications. What are the various parameters which correlate bond energy with bond strength?
Q2 (iv)
Write Limitations of Bohr’s theory. Which rule and principle is violated in writing the following E.C.:
➡️ 1s², 2s³ (Pauli’s exclusion principle; correct: 1s², 2s² 2pₓ¹)
➡️ 1s², 2pₓ² (Aufbau principle; correct: 1s², 2s²)
➡️ 1s², 2s², 2pₓ² 2pᵧ¹ (Hund’s rule; correct: 1s², 2s² 2pₓ¹ 2pᵧ¹ 2p_z¹)
➡️ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d⁴ 4s³ (Pauli’s exclusion principle and Hund’s rule; correct: [Ar], 3d⁵ 4s¹)
Q2 (v)
What is Hybridization? Give three differences between three types of hybridization. Draw the shape of ethene and ethyne molecules on the basis of Hybridization.
OR
Draw dot and cross structures of O₂, N₂, CH₄, CO₂, CHCl₃, C₂H₂, C₂H₄. Explain the ionic character of covalent bond.
(vi)
What are X-rays? How are they produced? Give their types and uses. What is the Relationship between Wavelength of X-rays and Nuclear Charge of Atom & how did Atomic Number discover by Moseley (role of X-rays in Moseley’s contribution)?
OR
Describe strength of covalent bond in term of VBT. Write down limitations of VBT.
(vii)
Define the process of hydrolysis. Explain the behavior of each of the following salts in aqueous solution with equations:
(a) K₂CO₃ (b) (NH₄)₂SO₄ (c) NaNO₃
OR
Define buffer, buffer action and buffer capacity? Explain the mechanism of buffer action with its applications.
(viii)
What do you mean by solubility product? Derive an expression for Ksp. Write down the solubility product expressions for the following sparingly soluble salts along with their units:
(i) Mg(OH)₂ (ii) Mg₃(PO₄)₂
What is meant by reaction rate? Enlist various factors which influence the rate of chemical reactions and describe the effect of temperature and surface area of solid reactant on reaction rate.
(ix)
What are the units for the rate constants for zero order, 1st order, 2nd order and 3rd order reactions?
OR
Differentiate between Elementary and Complex Reactions. What is reaction mechanism? Explain it with the help of example.
OR
Differentiate between homogeneous and heterogeneous catalysis. Give one example of each.
(x)
Define three major kinds of intermolecular forces in liquids. Explain intermolecular forces in HCl.
OR
What is London force? How is it originated? Describe factors on which strength of these forces depend.
OR
What is hydrogen bond? How is it established? Give its applications in industrial and biochemical processes. Describe hydrogen bonding in water and explain high specific heat, and the anomalous behaviour of water due to hydrogen bonding.
(xi)
What is oxidation number? State its any seven rules with examples. Find the oxidation number of the following (Write only value of oxidation numbers):
➡️ N in NH₄NO₃ ➡️ P in K₂MgP₄O₇ and Ca(H₂PO₄)₂ ➡️ S in Na₂S₄O₆ and Na₂S₂O₃
➡️ Ni in Ni(CO)₄ ➡️ Cr in Cr₂O₇²⁻ and H₂CrO₄ ➡️ Fe in Fe(CO)₃ and Fe₃O₄
➡️ Mn in MnO₄⁻ ➡️ C in CNO⁻ and C₃O₄ ➡️ P in POCl₃
➡️ Cl in CaOCl₂ ➡️ Cl in HOCl and HClO₄ ➡️ Br in BrO₃⁻
OR
What is meant by electrolysis? Explain the electrolysis of molten CaCl₂.
OR
Draw a fully labeled Born Haber cycle for Rubidium chloride (RbCl) and determine the lattice energy by using the following values (all in kJ/mol). (Answer: −692 kJ/mol)
➡️ I.P1st of Rb = 403 kJ/mol
➡️ Electron affinity of Cl = −349 kJ/mol
➡️ Bond energy of Cl₂ = 242 kJ/mol
➡️ Sublimation energy of Rb = 86.5 kJ/mol
➡️ Heat of formation of RbCl = −430.5 kJ/mol
(xii)
What are Colloids? Define their two types based on water as dispersion medium or physical state. Write down their Properties.
OR
What is liquid crystal? Give its two properties and two uses.
(xiii)
Define four types of solids according to bonding. Describe any two of them.
OR
Define lattice energy. Explain how it is affected by size and charge of ion?
OR
Define unit cell. How can you determine the number of Na⁺ and Cl⁻ ions in one unit cell of sodium chloride (NaCl)?
(xiv)
Define surface tension and viscosity with their units. Describe the two factors that affect them.
OR
What is the fourth state of matter? Give its significance in daily life.
(xv)
What is corrosion? What causes it to form? What can be done to prevent its formation?
(xvi)
State and explain Hess’s law of constant enthalpy summation. Calculate the enthalpy of combustion of propane at 25°C by the given information:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g) ∆H° = ?
∆Hf° of C₃H₈(g) = −103.9 kJ/mol
∆Hf° of CO₂(g) = −393.5 kJ/mol
∆Hf° of H₂O(g) = −285.8 kJ/mol
(xvii)
What are colligative properties? Why does the boiling point of a liquid get raised when a non‑volatile solute is added? Explain depression in freezing point of dilute solution.
OR
Define oxide and salt. Explain three types of oxides and four types of salts based on their acid‑base properties with examples.
(xviii)
Define Electrode potential and SHE. How is electrode potential of Zinc or Cu determined?
OR
What is battery? Define primary and secondary batteries. Describe the construction and working of lead storage battery with the help of relevant chemical equations or dry cell.
(xix)
Give Scientific reactions of any 4 of the following:
➡️ CO₂ is non‑polar while H₂O (or SO₂) is Polar molecule.
➡️ The boiling point of water is greater than that of HF even though F has greater EN than O.
➡️ s‑s sigma bond is weaker than s‑p or p‑p sigma bond.
➡️ 100 cm³ of O₂ and 100 cm³ of NH₃ contain the same number of molecules at STP.
➡️ The rates of diffusion of CO₂ and C₃H₈ gases are the same.
➡️ Evaporation is a cooling process.
➡️ Density of water is highest at 4°C.
➡️ Glycerin is distilled at reduced pressure.
➡️ H₂O forms concave meniscus while Hg forms convex meniscus.
➡️ Ethyl alcohol (C₂H₅OH) has greater viscosity than diethyl ether (C₂H₅OC₂H₅).
➡️ Powdered zinc or marble (CaCO₃) reacts quickly with hydrochloric acid than its solid lump.
➡️ Milk sours more rapidly in summer than in winter.
➡️ Why n‑hexane (petrol) is immiscible in water?
➡️ Pressure cooker is used for rapid cooking.
➡️ Vapour pressure of solution is lowered by adding non‑volatile solute.
➡️ A free falling drop of liquid is spherical.
➡️ Sigma bond is stronger than pi bond.
➡️ All photochemical reactions are zero order reactions.
➡️ Aqueous solution of Na₂CO₃ is alkaline while aqueous solution of NH₄Cl is acidic in nature.
➡️ Vapour pressure of solution is lowered by adding non‑volatile solute.
➡️ Solubility of oxygen in water increases with pressure but solubility of glucose in water has negligible effect of pressure.
(xx)
Do as directed (any 4 of the following):
➡️ Calculate wave number of an electron when it jumps from orbit 5 to orbit 2.
➡️ Arrange the following energy levels in ascending order using (n+l) rule: 5d, 3s, 4f, 7s, 6p, 2p
➡️ Define an electrochemical series? Give its properties.
➡️ Sketch a zinc‑copper Galvanic cell.
➡️ Define pH and pOH of a solution. Also show that pH + pOH = 14.
➡️ A solution is made by dissolving 14.8 g HCl in water at 25°C. If the volume of solution is 750 cm³ and HCl is assumed to be completely ionized, calculate its pH.
➡️ Calculate the molar mass of an unknown gas whose effusion rate is 2.83 times the effusion of methane.
➡️ The threshold energy of a reaction is 30 J. The average internal energy is 19 J. Calculate Ea.
➡️ Calculate the number of molecules and the volume in cm³ at STP of 0.28 g of nitrogen gas.
➡️ Calculate the density of oxygen gas at 45°C when the gas is confined in cylinder at 1170 torr pressure.
➡️ A cylinder contains 2.2 moles of oxygen gas at STP. When more oxygen gas is pumped into the cylinder, the volume of gas is changed from 2.0 dm³ to 3.4 dm³. Calculate how many moles of the oxygen gas are added to the cylinder? [Answer: n₂ = 3.74, moles added = 1.54 moles]
➡️ 150 cm³ of H₂ gas was collected over water at 800 mm Hg and 28°C. Calculate the mass in gram of H₂ gas obtained. The pressure of water vapours at 28°C is 20 mm Hg.
➡️ 40 g of NH₃ is confined in a vessel at STP. Find number of moles, number of molecules, volume of NH₃ gas and number of hydrogen atoms.
✍️ Section C
🔹 Attempt 4 Long Questions ➡️ 8 marks each 🎯 Total = 32 marks
⏰ Suggested Time for Solution
🕒 Total: 70 mins
✍️ Each part: 16 mins
Q3.
State the postulates of Bohr’s atomic theory with its defects? Derive an expression for the frequency and wave number expression of photon from E = −k (1/n²) or energy of nth stationary state of hydrogen atom?
OR
What are Rate Law and Rate Constant. Derive rate expression for a general chemical reaction. Enlist various factors which influence on the rate of chemical reactions and describe the effect of concentration of reactants, surface area and temperature on the reaction rate.
Q4.
State and explain first law of thermodynamics. Prove that:
(i) ∆H = qp by deriving Pressure volume work
(ii) ∆E = qv. Draw diagram where necessary.
OR
State the basic postulates of Kinetic Molecular Theory. State and explain Avogadro’s law or Charles law.
Q5.
State the basic postulates of VSEPR theory. Predict the geometry of molecules containing the following electron pairs on their central atom:
➡️ Three bond pairs and one lone pair (NH₃ or PH₃ or PCl₃)
➡️ Two bond pairs and two lone pairs (H₂O or H₂S or OF₂)
➡️ Three bond pairs and no lone pairs (AlCl₃ or BF₃)
➡️ Two bond pairs and no lone pair (CO₂ or CS₂)
➡️ Two bond pairs and one lone pair (SO₂)
OR
Explain the shapes of any three of the following molecules on the basis of hybridization and electron pair repulsion theory: Ethene, Ethyne, CH₄, H₂O, NH₃
OR
State the postulates of Molecular Orbital Theory. Draw the MO diagram of O₂/N₂ molecule. Find:
(i) Molecular orbital configuration
(ii) Bond order
(iii) Magnetic nature
Q6.
What are the conditions of deviation of real gas from ideal behaviour? Write down Graphical Explanation of Deviation of Real Gases. Derive van der Waal’s equation for real gases by correcting volume and pressure. Deduce Units for van der Waal’s Constants ‘a’ and ‘b’.
OR
Differentiate between ideal and non‑ideal gas. What are the conditions of deviation of real gas from ideal behaviour? Write down Graphical Explanation of Deviation of Real Gases. Discuss the deviation of Ideal behaviour of gases at low temperature and high pressure.
OR
Derive ideal gas equation by combining the gas laws. Calculate SI value and calorific value of molar gas constant.
Q7.
State and explain law of equilibrium. How can we predict the direction of reversible reaction with the help of reaction quotient? Derive Kc expression using this law for a general reaction and following reversible reaction: mN₂ + nH₂ ⇌ xNH₃.
State and explain Graham’s law or Dalton’s law of partial pressure with its applications.
Q8.
State Lewis concept or Bronsted‑Lowry concept of acids and bases with examples. Also explain conjugate acid‑base pair. Write equation and indicate the conjugate acid‑base pairs for the following:
(i) Acetic acid and ammonia
(ii) Hydrochloric acid and water
(iii) Ammonia and water
Q9.
State Le‑Chatelier’s Principle. Describe the effect of change in concentration and change in pressure on Equilibrium. Explain the industrial application of Le‑Chatelier’s principle using Haber’s process.
Predict the effect of increase in temperature and pressure on the following systems at equilibrium state (only predict the direction):
(i) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + Heat
(ii) N₂(g) + O₂(g) + Heat ⇌ 2NO(g)
Q10.
State Raoult’s law. Derive this law mathematically in three forms. Under what conditions does Raoult’s law apply?
OR
Define osmosis and semipermeable membrane. Explain that osmosis is a colligative property of solution. Give four daily life examples of osmosis.
Q11.
What is redox reaction? Balance any two of the following equations by ion‑electron method (No description is required):
➡️ Cr(OH)₃ + H₂O₂ → CrO₄²⁻ + OH⁻ (basic medium)
➡️ Cl₂ + OH⁻ → Cl⁻ + ClO₃⁻ + H₂O (basic medium)
➡️ H₂S + HNO₃ → S + NO + H₂O (acidic medium)
➡️ MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻ (acidic medium)
✏️ ✨📊 Key Numericals 🔢 🎯 ✏️
🌟 Chapter # 1
Q1.
How many gram of NH₃ are formed when 100 g of each of the following reagents are reacted together according to following equation:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O
(Answer: 31.773 g NH₃)
How many grams of sodium thiosulphate (Na₂S₂O₃) will be produced when 200 g of each of Na₂S, Na₂CO₃ and SO₂ are reacted together.
2Na₂S + Na₂CO₃ + 4SO₂ → 3Na₂S₂O₃ + CO₂
(Answer: 369.72 g Na₂S₂O₃)
Hydrogen gas is commercially prepared by steam methane process:
CH₄ + H₂O → CO + 3H₂
If a mixture of 28.8 g methane and 14.4 g steam is heated in a furnace, determine the limiting reactant and the mass and volume of hydrogen gas produced.
(Answer: H₂O is LR, moles of H₂ = 2.4, Mass = 4.8 g, Volume = 53.76 dm³)
60 g of hydrogen sulphide (H₂S) burns with 40 g oxygen (O₂):
2H₂S + 3O₂ → 2SO₂ + 2H₂O
Calculate the mass and volume of SO₂ gas at STP.
(Answer: 60 g SO₂, Volume = 42 L)
Under high pressure magnesium reacts with oxygen:
2Mg + O₂ → 2MgO
If 4 g of Mg produces 4.24 g of MgO, calculate percentage yield.
(Answer: 63.85%)
6.8 g of Mg reacts with excess of O₂ to produce 10.44 g of MgO.
Calculate the percentage yield.
(Answer: 95.6%)
Q2.
Aluminium Sulphide is prepared by the reaction of aluminium metal and sulphur powder at elevated temperature:
2Al + 3S → Al₂S₃
If 135 g aluminium and 160 g sulphur are taken for the reaction, calculate what mass of Al₂S₃ will be formed. If 155 g of Al₂S₃ is obtained then find out the % yield of this reaction.
(Answer; Theoretical yield = 222.2 g Al₂S₃, % yield = 69.8%)
OR
When aluminium is heated with nitrogen at 700°C, it gives aluminium nitride:
2Al(s) + N₂(g) → 2AlN(s)
If 67.5 g of aluminium and 140 g of nitrogen gas are allowed to react, find out:
(a) Limiting reactant and Mass of aluminum nitride produced?
(b) Mass of excess reactant?
(Answer; Al is LR, 102.5 g AlN produced, 105 g N₂ left behind)
🌟 Chapter # 2
Q3.*
A photon of wave number 23 × 10⁵ m⁻¹ is emitted when electron undergoes a transition from a higher orbit to n = 2. Determine the orbit from which electron falls and also the spectral line appears in this transition of electron (The value of Rydberg constant is 1.09678 × 10⁷ m⁻¹).
[Ans; n₂ = 5.05 ≈ 5, Number of spectral lines = 6]
🌟 Chapter # 4
Q4.
Four containers of equal volume are filled as follows: [Hint: Suppose Volume = 1 dm³].
(i) 2.0 g of H₂ at 0°C (ii) 1.0 g of H₂ at 273°C
(iii) 24 g of O₂ at 0°C (iv) 16 g of CH₄ at 273°C
(a) Which container is at the greatest pressure?
(b) Which container is at the lowest pressure?
[Answer; Container (iv) is at highest pressure, Container (iii) is at lowest pressure]
(Ans: P₁ = 22.4 atm; P₂ = 22.4 atm; P₃ = 16.8 atm; P₄ = 44.8 atm)
Q5.
40 dm³ of hydrogen gas was collected over water at 831 torr pressure at 23°C. What would be the volume of dry hydrogen gas at standard conditions? The vapour pressure of water at 23°C is 21 torr of Hg.
[Answer; V₂ = 39.32 dm³]
Q6.
A 500 cm³ vessel contains H₂ gas at 400 torr pressure and another 1 dm³ vessel contains O₂ gas at 600 torr pressure. If under the similar condition of temperature these gases are transferred to 2 dm³ empty vessel, calculate the pressure of the mixture of gases in new vessel.
[Answer; PT = 100 + 300 = 400 torr]
Q7.
Two moles of oxygen gas is kept in a vessel of 15.5 dm³ at a temperature of 37°C. Calculate the pressure exerted by the gas if:
(a) gas behaves as ideal
(b) gas behaves as non-ideal
a = 1.36 atm dm⁶ mol⁻² and b = 0.0318 dm³ mol⁻¹.
[Answer; Pideal = 3.28 atm, Preal = 3.27 atm]
Q8.
At a specific temperature and pressure, it takes 290 s for a 1.5 dm³ sample of He to effuse through a porous membrane. Under similar conditions, if 1.5 dm³ of an unknown gas “X” takes 1085 s to effuse, calculate the molar mass of gas “X”.
[Mx = 55.96 ≈ 56 g/mol]
Q9.
If it takes 8.5 seconds for 5 cm³ of CO₂ gas to effuse through a porous material at a particular temperature and pressure. How long would it take for 5 cm³ of SO₂ gas to effuse from the same container at the same temperature and pressure?
[Answer; 10.43 seconds]
Q10.
Oxygen gas is produced by heating potassium nitrate:
2KNO₃ → 2KNO₂ + O₂
The gas is collected over water. If 225 cm³ of gas is collected at 25°C and 785 mm Hg total pressure, what is the mass of O₂ gas collected? (Pressure of vapours at 25°C is 23.8 mm Hg)
[Answer; nO₂ = 9.2 × 10⁻³ mole, Mass = 0.294 g O₂]
Q11.
A chemist has synthesized a gas whose density was found to be 1.88 g dm⁻³ at 27°C and 760 torr pressure. Calculate its molar mass.
[Answer; M = 46.0 g/mol]
Q12.
The volume of hydrogen gas collected over at 24°C and 762 mmHg pressure is 1280 cm³. Calculate mass of hydrogen gas in gram obtained. The aqueous tension of water at 24°C is 22 mmHg.
The rate of diffusion of an unknown gas is 70.3 cm³/s whereas CO₂ is 60 cm³/s under similar conditions. What is the molecular mass and identity of the unknown gas?
[Answer; Mass of H₂ = 0.101 g; Molar mass of unknown gas ≈ 44 g/mol, Identity = CO₂]
🌟 Chapter # 7
Q1.
Following reaction was studied at 25°C:
2NO(g) + Cl₂(g) ⇌ 2NOCl(g)
At equilibrium, the partial pressures are:
PNOCl = 1.2 atm,
PNO = 5 × 10⁻² atm,
PCl₂ = 3 × 10⁻¹ atm.
Calculate the values of Kp and Kc.
[Answer; Kp = 1.92 × 10³ atm⁻¹, Kc = 1.89 × 10³ mol⁻¹ dm³]
Q2.
Nitrosyl chloride is a yellow coloured gas prepared by the reaction of NO and Cl₂ gases:
2NO(g) + Cl₂(g) ⇌ 2NOCl(g)
If at certain temperature, the partial pressure of equilibrium mixture is:
NO = 0.17 atm, Cl₂ = 0.2 atm and NOCl = 1.4 atm, calculate Kp.
(Answer; 339.1 atm⁻¹)
Q3.
Ethyl acetate is an ester of ethanol and acetic acid commonly used as an organic solvent:
CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
In an esterification process, 180 g of acetic acid and 138 g of ethanol were mixed at 298K and allowed to start reaction under necessary conditions. After equilibrium is established, 60 g of unused acid were present in the reaction mixture. Calculate Kc.
(Answer; 4)
OR
At 444°C reaction of hydrogen and iodine is performed in a sealed 1 dm³ steel vessel:
H₂(g) + I₂(g) ⇌ 2HI(g)
If equilibrium mixture contains 1 mole of H₂, 1 mole of I₂ and 7 moles of HI, calculate:
(a) Equilibrium constant (Kc)
(b) Initial concentration of H₂ and I₂
(Answer; [H₂]initial = 4.5 mol/dm³, [I₂]initial = 4.5 mol/dm³)
Q4.
Calculate the value of Kₚ for the given reaction:
N₂ + 3H₂ ⇌ 2NH₃ at 27°C if the value of Kᴄ is 0.0012 mol⁻² dm⁶. (R = 0.0821 atm·dm³·mol⁻¹·K⁻¹)
[Answer; Kₚ = 2.99 × 10⁻⁵ atm⁻²]
Q5.
The reaction of methane with hydrogen sulphide gives carbon disulphide:
CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g)
If Kc for this reaction at 727°C is 4.2 × 10⁻³, calculate its Kₚ.
(Answer; 28.30)
Q6.
Kc for the given reaction at certain temperature is 2.72:
A(g) + 3B(g) ⇌ 2C(g)
If in a 5 dm³ vessel the reaction mixture contains 8 moles A, 6 moles B and 5 moles C. Predict whether the reaction is in equilibrium, if not what is the direction of net reaction.
(Answer; Q = 0.36 dm³ mol⁻¹, Forward Reaction)
Q7.
164 g H₂ and 518.4 g I₂ are mixed and reacted in a sealed 1 dm³ steel vessel at 444°C until the equilibrium is established, 338 g of HI is formed. Calculate Kᴄ.
H₂ + I₂ ⇌ 2HI
[Answer; Kᴄ ≈ 49]
Q8.
A solution is prepared by mixing 600 cm³ of 7.5 × 10⁻⁴ M BaCl₂ into 300 cm³ of 2.4 × 10⁻³ M Na₂SO₄ solutions. Will precipitate of BaSO₄ form when equilibrium is reached? (Kₛₚ of BaSO₄ = 1.1 × 10⁻¹⁰ mol² dm⁻⁶)
[Answer; Q = 3.6 × 10⁻⁷ > Kₛₚ, Precipitate forms]
Q9.
The solubility of AgCl at 25°C is 1.4 × 10⁻³ g/dm³. Its molecular mass is 143.5. Calculate molarity of AgCl solution and solubility product of AgCl.
(Kₛₚ = 9.5 × 10⁻¹¹ M²)
Q10.
Find the solubility of MgF₂ when its Kₛₚ is 7.26 × 10⁻⁹ mol³/dm⁹.
(x = 1.219 × 10⁻³ mol/dm³, Solubility of MgF₂ = 7.55 × 10⁻² g/dm³)
Q11.
Should AgCl precipitate from a solution prepared by mixing 400 cm³ of 0.1 M NaCl and 600 cm³ of 0.03 M AgNO₃. Kₛₚ of AgCl is 1.6 × 10⁻¹⁰ mol²/dm³.
(Q = 7.2 × 10⁻⁴ M², AgCl should form ppt)
🌟 Chapter # 9
Q1.
Decomposition of NO₂ into NO and O₂ is of second order reaction:
2NO₂ → 2NO + O₂
If the rate constant at certain temperature is 3.8 × 10⁻⁴ dm³ mol⁻¹ s⁻¹ and the initial concentration of NO₂ is 0.38 M, calculate the initial rate of reaction.
[Answer; Rate = k[NO₂]² = 3.8 × 10⁻⁴ × (0.38)² = 5.49 × 10⁻⁵ mol dm⁻³ s⁻¹]
Q2.
The rate constant for the decomposition of NOCl is 2.48 × 10⁻² dm³ mol⁻¹ s⁻¹. Calculate the initial rate when initial concentration of reactant is 0.65 mol/dm³. What will happen to the rate of reaction and rate constant, if the concentration of NOCl is doubled?
2NOCl → 2NO + Cl₂
[Answer; Rate = k[NOCl]² = 2.48 × 10⁻² × (0.65)² = 0.0105 mol dm⁻³ s⁻¹. If [NOCl] is doubled, rate becomes 4× (0.042 mol dm⁻³ s⁻¹), but k remains constant.]
Q3.
The rate constant (k) for the decomposition of NO₂ is 1.8 × 10⁻³ dm³ mol⁻¹ s⁻¹.
2NO₂ → 2NO + O₂
Write the rate expression.
Find the initial rate when the initial concentration of NO₂ is 0.75 mol dm⁻³.
Find the value of the rate constant when the initial concentration of NO₂ is doubled.
[Answer; Rate expression: Rate = k[NO₂]². Initial rate = 1.8 × 10⁻³ × (0.75)² = 1.01 × 10⁻³ mol dm⁻³ s⁻¹. If [NO₂] is doubled (1.5 M), rate = 1.8 × 10⁻³ × (1.5)² = 4.05 × 10⁻³ mol dm⁻³ s⁻¹. Rate constant k remains unchanged = 1.8 × 10⁻³ dm³ mol⁻¹ s⁻¹.]
🌟 Chapter # 9
Q4.
The initial rate data in a series of experiments while working on the hydrolysis of ester into acetic acid and ethyl alcohol is given in the following table. Determine its rate law, order of reaction and rate constant along with unit.
CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
| Experiment # | Initial [CH₃COOC₂H₅]; M | Initial [H₂O]; M | Initial rate; M s⁻¹ |
|---|---|---|---|
| 1 | 0.1 | 0.1 | 4 × 10⁻⁴ |
| 2 | 0.4 | 0.1 | 1.6 × 10⁻³ |
| 3 | 0.1 | 0.4 | 4 × 10⁻⁴ |
OR
The initial rate data in a series of experiments while working on the oxidation of nitric oxide to give nitrogen dioxide is given in the following table. Determine its rate law and find the order of reaction.
2NO + O₂ → 2NO₂
| Experiment # | Initial [NO]; M | Initial [O₂]; M | Initial rate; M s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 8 × 10⁻⁴ |
| 2 | 0.10 | 0.20 | 16 × 10⁻⁴ |
| 3 | 0.20 | 0.10 | 16 × 10⁻⁴ |
OR
Determine the order of reaction and rate constant by the given data:
| Initial [A]; mol dm⁻³ | Initial [B]; mol dm⁻³ | Initial rate; mol dm⁻³ s⁻¹ |
|---|---|---|
| 0.1 | 0.1 | 2.52 × 10⁻³ |
| 0.2 | 0.1 | 10.08 × 10⁻³ |
| 0.1 | 0.2 | 5.04 × 10⁻³ |
OR
The experimental data for the kinetic study of a reaction; A + B → C, at 80°C is given below:
Determine the order of reaction and rate constant by the given data:
| Experiment # | Initial [A]; mol dm⁻³ | Initial [B]; mol dm⁻³ | Initial rate of C; mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.1 | 0.05 | 1.25 × 10⁻⁴ |
| 2 | 0.2 | 0.05 | 2.5 × 10⁻⁴ |
| 3 | 0.1 | 0.1 | 5.0 × 10⁻⁴ |
🌟 Chapter # 10
Q1.
Glucose is a non-volatile solute in water. A glucose solution contains 0.15 moles glucose and 5.8 moles water at 20°C. Determine the lowering in the vapour pressure if the vapour pressure of pure water at 20°C is 17.5 torr. (Assume solution is an ideal).
(Answer; ∆P = P° X₂ = 17.5 × 0.0252 = 0.441 torr)
Q2.
Calculate the molality of a 12% urea solution (molar mass of urea is 60 g/mol).
(Answer; 2.27 mol/kg)
Q3.
Define molality of solution. Calculate the molality of 25% (w/w) aqueous solution of glucose (molar mass of glucose is 180 g/mol).
(Answer; Molality = 1.85 mol/kg)
Q4.
An aqueous solution of 1.2 molality is prepared by dissolving some amount of oxalic acid into 475 g water. Determine the mass of oxalic acid in the solution. (Molecular mass of oxalic acid is 126 g/mol).
(Answer; 71.82 g)
Q5.
45 g glucose dissolves in 72 g water to make a solution. Calculate the mole fraction of glucose and water in the solution.
(Answer; 0.0588, 0.941)
Q6.
In the analysis of water sample, it was reported that 1 g of water contains 6.34 × 10⁻³ mg magnesium ions. Calculate the concentration of magnesium ion in ppm.
(Answer; 6.34)
Q7.
Automotive antifreeze is a 60% aqueous solution of ethylene glycol (C₂H₆O₂). Determine:
(a) molality of solution
(b) mole fraction of ethylene glycol in the solution
(Answer; molality = 24.19 mol/kg, mole fraction of ethylene glycol = 0.3, mole fraction of water = 0.693)
Q8.
The vapour pressure of a pure liquid A is 37 mm Hg at 27°C. It is mixed into another liquid B to make a solution. The vapour pressure of A in the solution is found to be 33 mm Hg. Calculate the mole fraction of A (Assume it obeys Raoult’s law).
(Answer; 0.89)
🌟 Chapter # 11
Q1.
Calculate the standard enthalpy of formation of Glycerol from the data given below.
➡️ 3C₍ₛ₎ + 4H₂₍g₎ + 3/2O₂ → C₃H₈O₃₍ₗ₎ ΔHf° = ?
➡️ C₍ₛ₎ + O₂₍g₎ → CO₂₍g₎ ΔH° = −393.5 kJ/mol
➡️ H₂(g) + ½O₂₍g₎ → H₂O₍ₗ₎ ΔH° = −285.8 kJ/mol
➡️ C₃H₈O₃₍ₗ₎ + 3½O₂(g) → 3CO₂(g) + 4H₂O(l) ΔH° = −1654.1 kJ/mol
[Answer; ΔHf°(C₃H₈O₃) = −669.5 kJ/mol]
Q2.
Calculate the heat of formation of ethane at 25°C from the following data:
➡️ 2C₍ₛ₎ + 3H₂₍g₎ → C₂H₆₍g₎ ΔHf° = ?
➡️ C₍ₛ₎ + O₂₍g₎ → CO₂₍g₎ ΔH = −394 kJ/mol
➡️ H₂₍g₎ + ½O₂₍g₎ → H₂O₍g₎ ΔH = −286 kJ/mol
➡️ C₂H₆ + 7/2 O₂₍g₎ → 2CO₂₍g₎ + 3H₂O₍g₎ ΔH = −1561 kJ/mol
[Answer; ΔHf°(C₂H₆) = −85 kJ/mol]
Q3.
A thermochemical process is carried out at constant pressure of 8.52 atm. If it absorbs 15.5 kJ energy from the surrounding, due to which an expansion in the volume of 4.7 dm³ occurs, calculate its change in internal energy.
[Answer; ΔE = q − PΔV = 15.5 − (8.52 × 4.7 × 0.1013) = 11.44 kJ]
OR
A thermochemical process is carried out at constant pressure of 2759 N/m². If it absorbs 15400 J heat energy from the surrounding and the volume increases by 0.47 m³ by pushing the piston upward, calculate the change in internal energy in kJ.
[Answer; ΔE = 15400 − (2759 × 0.47) = 14100 J ≈ 14.1 kJ]
📝 Multiple Choice Questions (MCQs) – Compulsory Section (17 Marks) 🎯📌
1. The total number of ions in CaCl₂ is
Reason: 1 mole CaCl₂ gives 3 ions. Total ions = 3 × 6.02×10²³ = 18.06×10²³.
2. Which one of the following cannot have value equal to zero?
Reason: Activation energy cannot be zero for most reactions, and principal quantum number starts from 1, never zero.
3. If the volume occupied by oxygen gas (O₂) at STP is 44.8 dm³, the number molecules of O₂ in the vessel are:
Reason: 22.4 dm³ at STP = 1 mole. 44.8 dm³ = 2 moles. Molecules = 2 × 6.02×10²³ = 12.04×10²³.
4. The number of carbon atoms in half mole of sugar (C₁₂H₂₂O₁₁) are approximately:
Reason: Half mole sugar = 0.5×6.02×10²³ molecules. Each molecule has 12 C atoms. Total C atoms ≈ 36×10²³.
5. Rate = K[NH₃]². Keeping the other conditions same, if the concentration of NH₃ is doubled, then the initial rate of reaction will be
Reason: Rate ∝ [NH₃]². Doubling [NH₃] increases rate by 2² = 4 times.
6. The n+l value for 4d orbital is
Reason: For 4d, n=4, l=2. So n+l = 6. Correction: Actually 4+2=6. Correct Answer: (d) 6.
7. Which of the following possesses weakest London dispersion forces:
Reason: Smaller molecules have weaker dispersion forces. F₂ is smallest, so weakest forces.
8. In NaCl, each Na ion is surrounded by Cl ions in the numbers:
Reason: NaCl has octahedral geometry; each Na⁺ is surrounded by 6 Cl⁻ ions.
9. This molecule has zero dipole moment
Reason: CCl₄ (tetrahedral) and CO₂ (linear) are symmetrical, so dipole moments cancel.
10. The number of bonds in ethyne (C₂H₂) is:
Reason: Ethyne has a triple bond (1 sigma + 2 pi) and one sigma bond between carbons and hydrogens. Total = 3 sigma, 2 pi.
11. The potential energy of an electron can be denoted by
Reason: In Bohr’s model, potential energy of electron = −(Ze²)/(4πε₀r).
12. A colloidal solution of liquid into liquid is known as:
Reason: Emulsion is a colloidal system of liquid dispersed in liquid.
14. The unit of rate constant (K) for the first order reaction is:
Reason: For first order reactions, rate constant has units of reciprocal time.
15. The decomposition of H₂O₂ is inhibited by:
Reason: Glycerin stabilizes H₂O₂ and prevents its decomposition.
16. This molecule has the maximum bond angle
Reason: CO₂ is linear with bond angle 180°, maximum among the given molecules.
17. The number of orbitals in each energy level is given by the formula
Reason: Each energy level has n² orbitals.
18. Salt which is formed by the neutralization of weak acid and strong base is:
Reason: Na₂CO₃ is formed from weak acid H₂CO₃ and strong base NaOH.
19. The number of bonds in ethene (C₂H₄) is:
Reason: Ethene has 4 C–H sigma bonds, 1 C–C sigma bond, and 1 C–C pi bond.
20. If the volume occupied by oxygen gas (O₂) at STP is 44.8 dm³, the number of atoms of O₂ in the vessel are:
Reason: 44.8 dm³ = 2 moles O₂ molecules = 12.04×10²³ molecules. Each molecule has 2 atoms → 24.08×10²³ atoms.
21. The number of carbon atoms in one mole of sugar (C₁₂H₂₂O₁₁) are approximately:
Reason: 1 mole sugar = 6.02×10²³ molecules. Each molecule has 12 C atoms → 72×10²³ atoms.
22. The strongest oxidizing agent and strongest reducing agent in the electrochemical series respectively is:
Reason: Li is strongest reducing agent, F is strongest oxidizing agent.
23. Least entropy found in which of the following state of water:
Reason: Solid state has least entropy compared to liquid or gas.
24. Which of the following possesses strongest London dispersion forces:
Reason: Larger molecules have stronger dispersion forces. I₂ is largest, so strongest.
25. In one unit cell of sodium chloride (NaCl), the total number of sodium and chloride ions is:
Reason: NaCl unit cell has 4 formula units, i.e. 4 Na⁺ and 4 Cl⁻ ions.
26. The kinetic energy of an electron can be denoted by:
Reason: In Bohr’s model, kinetic energy = −(total energy) = Ze²/(8πε₀r).
27. Which of the following does NOT alter the pH of a solution?
Reason: NaCl is a neutral salt (strong acid + strong base), so does not alter pH.
28. The unit of rate constant (K) for the second order reaction is:
Reason: For second order reactions, rate constant has units of reciprocal concentration times reciprocal time.
29. The sum of mole fractions of components of a solution is equal to:
Reason: By definition, sum of mole fractions of all components in a solution = 1.
30. The conduction of electricity through an electrolytic solution is due to the flow of:
Reason: Electrolytic conduction occurs due to movement of cations and anions in solution.
📝 Most Important MCQs for XI Chemistry K.B 2026🎯📌
Reason: For a 3d orbital, the main shell (n) is 3, and for d-subshells, the azimuthal quantum number (l) is always 2.
Reason: Four times — because V∝T/P, doubling T and halving P makes V four times.
Reason: Entropy measures disorder. Gases (steam) have much higher molecular disorder than liquids or solids.
Reason: Lattice energy is the energy released when gaseous ions combine to form one mole of an ionic solid crystal.
Reason: An aerosol is a colloid where solid particles or liquid droplets are dispersed in a gas (e.g., smoke or dust in air).
Reason: Water is polar and Benzene is non-polar; they do not mix at all. Phenol/Water is partially miscible, while the others are completely miscible.
Reason: % W/W = (Mass of solute / Total mass of solution) × 100. For a 15% solution, 15g solute + 85g solvent = 100g total solution.
Reason: Combustion is an exothermic process because it releases energy in the form of heat. The others require energy input to proceed.
Reason: An isolated system is one that cannot exchange either matter or energy (including heat) with its surroundings.
Reason: Constant volume — because work W=PΔV, and if volume doesn’t change, no work is done.
Reason: Internal energy (ΔE) is a state function, meaning it only depends on the initial and final states. Heat (q) and Work (w) are path functions.
Reason: In a standard dry cell (Leclanché cell), the zinc container serves as the negative electrode (anode) and is oxidized during the reaction.
Reason: Metallic conduction occurs through the movement of "delocalized" or free electrons within the crystal lattice.
Reason: Fluorine has the highest reduction potential in the electrochemical series, making it the most powerful oxidizing agent (it gains electrons very easily).
Reason: Oxidation always occurs at the anode. Since Zinc is more reactive than Hydrogen, it loses electrons to become Zn²⁺.
Reason: "Increases by square" implies 2nd order for x (2²=4). "Increases twice" implies 1st order for y (2¹=2). Hence, R = K[x]²[y]¹.
Reason: For zero order, Rate = K. Since rate is concentration/time, the unit is molL⁻¹s⁻¹ (conc. s⁻¹).
Reason: In zero-order reactions, the rate is independent of the concentration of reactants, so the rate law is simply R = K[A]⁰, which is R = K.
Reason: Ionic reactions involve the simple combination of oppositely charged ions in solution, which happens almost instantaneously (Fast reactions).
Reason: The "velocity" of a reaction at a specific moment in time is most accurately described as the instantaneous rate.
Reason: The negative sign indicates that the concentration of reactant 'A' is decreasing over time.
Reason: NaOH (strong base) and HCl (strong acid) neutralize each other to form a salt and water; they do not form a buffer system, which requires a weak component.
Reason: Aluminum oxide (Al₂O₃) can react with both acids and bases, making it amphoteric.
Reason: KNO₃ is a salt of a strong acid (HNO₃) and a strong base (KOH). It does not undergo hydrolysis and remains neutral (pH 7).
Reason: According to Brønsted-Lowry theory, a conjugate acid is formed when a base (NH₃) accepts a proton (H⁺).
Reason: NaNO₃ is formed from NaOH (strong base) and HNO₃ (strong acid).
Reason: NH₄CN is formed from NH₄OH (weak base) and HCN (weak acid).
Reason: Rate ratio = √(Molar mass of O₂ / Molar mass of H₂) = √(32 / 2) = √16 = 4. So, the ratio is 4:1.
Reason: According to the Ideal Gas Law (V = nRT/P), volume is maximum when temperature (T) is highest and pressure (P) is lowest.
Reason: Rate ratio = √(Molar mass of SO₂ / Molar mass of CH₄) = √(64 / 16) = √4 = 2.
Reason: London dispersion forces increase with molecular size and number of electrons. F₂ is the smallest molecule in this group, so it has the weakest forces.
Reason: Non-polar molecules only experience London dispersion forces. As size increases, these forces become stronger due to increased polarizability.
Reason: Lattice energy is the energy released when gaseous ions combine to form one mole of an ionic solid.
Reason: Anisotropy is the variation of physical properties (like electrical conductivity) depending on the direction of measurement.
Reason: Transition temperature is the specific temperature where one allotrope changes into another and both exist in equilibrium.
Reason: Diamond consists of carbon atoms bonded together in a continuous 3D network of covalent bonds.
Reason: In ice, water molecules are held together by strong hydrogen bonds, creating a hexagonal open-cage structure.
Reason: In the Orthorhombic system,a≠b≠c
α=β=γ=90∘
Reason: Alum (Potassium aluminum sulfate) forms crystals with a regular internal arrangement. Glass, plastic, and rubber are amorphous solids.
Reason: A bond angle of 104.5° (like in H₂O) indicates a Bent or V-shaped geometry due to the presence of two lone pairs.
Reason: CS₂ is a linear molecule (like CO₂) with a bond angle of 180°. BF₃ is 120°, NH₃ is 107.5°, and H₂O is 104.5°.
Reason: Boron has 3 valence electrons and forms 3 bond pairs with no lone pairs, resulting in sp² hybridization and a planar trigonal shape.
Reason: Ammonia (NH₃) has sp³ hybridization with 3 bond pairs and 1 lone pair, which pushes the bonds down into a pyramidal shape.
Reason: With 5 electron pairs (2 bp + 3 lp), the arrangement is trigonal bipyramidal. To minimize repulsion, the 3 lone pairs occupy equatorial positions, leaving the atoms in a linear 180° arrangement (e.g., XeF₂).
Reason: 1 Debye is defined as 3.33564 x 10⁻³⁰ Coulomb-meters in SI units.
Reason: Nitrogen atoms share three pairs of electrons to form a triple bond (N≡N), so the bond order is 3.
Reason: In Ethene (CH₃-CH₃), there are 6 C-H sigma bonds, 1 C-C sigma bond (total 5 σ), and no C-C pi bond (0 π).
Reason: Benzene is a planar hexagonal ring where each carbon is sp² hybridized, resulting in 120° bond angles, same as trigonal planar (triangular) geometry.
Reason: In BF₃, SO₃, and CH₃⁺, the central atoms all have 3 electron domains and are sp² hybridized. (Note: Most papers accept 'a' or 'c', but both are technically sp²).
Reason: Each Oxygen in O=O has 1 sigma bond and 2 lone pairs (3 domains), which corresponds to sp² hybridization.
Reason: sp³ orbitals have the least 's' character (25%) and the most 'p' character (75%), making them the longest and most diffused hybrid orbitals. Therefore, sp³-sp³ overlapping results in the longest bond.
Reason: Oxygen has 2 bond pairs (with F) and 2 lone pairs. Each Fluorine has 3 lone pairs. Total lone pairs = 2 (on O) + 6 (on two F) = 8. Bond pairs = 2.
Reason: Both CO₂ and BeF₂ are linear molecules with a bond angle of 180°. H₂S, SnCl₂, and SO₂ all have bent/angular geometries.
Reason: Each water molecule can form 4 hydrogen bonds: two through its hydrogen atoms and two through the lone pairs on the oxygen atom.
Reason: According to the Pauli Exclusion Principle, a single orbital (atomic or molecular) can hold a maximum of 2 electrons with opposite spins.
Reason: Hund's Rule states that orbitals of equal energy are each occupied by one electron before any orbital is doubly occupied.
Reason: Each element has a unique line spectrum (fingerprint), which is used in spectroscopy to identify elements in a sample.
Reason: Isoelectronic species have the same number of electrons. Na⁺, Ne, F⁻, and Mg²⁺ all have 10 electrons. Na has 11, while Ca has 20.
Reason: Transitions to the 1st orbit (n=1) form the Lyman series, while transitions to the 2nd orbit (n=2) form the Balmer series.
Reason: When the azimuthal quantum number l = 0, it represents an 's' orbital, which is always spherical in shape.
Reason: At STP, 22.4 dm³ = 1 mole. So, 44.8 dm³ = 2 moles of O₂ molecules. Since each O₂ molecule has 2 atoms, total atoms = 2 moles x 2 atoms/molecule x 6.02 x 10²³ = 24.08 x 10²³.
Reason: 1 mole of sugar contains 12 moles of Carbon atoms. Total atoms = 12 x 6.02 x 10²³ = 72 x10²³.
Reason: Molar mass of Na = 23g. 92g Na = 4 moles. From the equation, 2 moles Na produce 1 mole H₂. Thus, 4 moles Na produce 2 moles H₂. Volume = 4 x 22.4 = 44.8 dm³.
Reason: 6g C is 0.5 moles (6/12). 16g S is 0.5 moles (16/32). 20g Ca is 0.5 moles (20/40). 12g Mg is 0.5 moles (12/24). They all contain the same number of atoms.
Reason: Moles of C = 60 / 12 = 5 moles. Number of atoms = 5 x 6.02 x10²³ = 30.1 x 10²³.
Reason: For a fixed mass, the substance with the lowest molar mass has the highest number of moles (and molecules). CH₄ (16 g/mol) is the lightest among these.
Reason: Mercury has extremely strong metallic bonding, giving it a much higher surface tension (approx 485 mN/m) than water (72 mN/m).
Reason: Radius ratio = r⁺/r- = 0.74 / 1.84 = 0.402.
Reason: Milk is a liquid-in-liquid colloid (fat globules dispersed in water), which is known as an emulsion.
Reason: H₂SO₄ is diprotic. [H⁺] = 2 x 0.001 = 0.002 M. pH = -log(0.002)= 2.69.
Reason: Molar mass of N₂ = 28 g/mol. Moles = 7/28 = 0.25 moles. Volume at STP = 0.25 x 22.4 = 5.6 dm³.
Reason: Sodium acetate (CH₃COONa) is a salt of a weak acid and a strong base. It undergoes anionic hydrolysis, producing OH⁻ ions, making the solution basic (pH > 7).
Reason: According to Graham's Law, the rate of diffusion is inversely proportional to the square root of the molar mass (r ∝ √ molar mass(M)). Since N₂ has the lowest molar mass (28) among the choices, it diffuses fastest.
Reason: Since 1 calorie = 4.184 Joules, then 1 Joule = 1 / 4.184 = 0.239 calories.
Reason: Heat (q) and Work (w) depend on the path taken between states, whereas Enthalpy is a state function that depends only on the initial and final states.
Reason: Options (a), (b), and (c) form near-ideal solutions. Ethanol and Acetone show a positive deviation from Raoult's Law because the hydrogen bonding in ethanol is disrupted by acetone.
Reason: According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Pressure has negligible effect on the solubility of solids and liquids.
Reason: The relative lowering of vapour pressure is equal to the mole fraction of the solute.
Reason: Entropy measures disorder. Solids (ice) have much higher order and lower molecular motion than liquids or gases, resulting in the lowest entropy.
Reason: London dispersion forces increase with the size and number of electrons (polarizability). Neon (Ne) is the smallest atom with the fewest electrons in this list, hence it has the weakest forces.
Reason: While VBT explains bond formation via orbital overlap, the specific concept of "Bond Order" (calculated as (Nb - Na)/2) is a primary feature of Molecular Orbital Theory (MOT), not VBT.
Reason: Non-polar molecules do not have permanent dipoles, so they only experience London dispersion forces. These forces become stronger as the molecule's size (and polarizability) increases.
Reason: Combustion is an exothermic process, meaning energy is always released (∆H < 0). Fusion and atomization are endothermic (∆H > 0), and solution can be either.
Reason: The Joule-Thomson effect describes the temperature change of a real gas when it is allowed to expand freely through a valve or porous plug. This cooling effect is used to liquefy gases.
Reason: 6.02 x 10²³ molecules (1 mole) occupy 22.4 dm³. Therefore, 3.01 x 10²³ molecules (0.5 mole) occupy 22.4 / 2 = 11.2 dm³.
Reason: 1 mole of H₂O has 2 moles of Hydrogen atoms. Total atoms = 2 x 6.02 x 10²³ = 12.04 x 10²³ = 1.204 \times 10²⁴.
Reason: Moles of H₂ = 10/2 = 5. Moles of CH₄ = 80/16 = 5. Since the number of moles is the same, the number of molecules is the same.
Reason: At the boiling point (100°C for water), the vapor pressure of a liquid equals the atmospheric pressure, which is 760 torr (or 1 atm).
Reason: There are 7 basic crystal systems: Cubic, Tetragonal, Orthorhombic, Rhombohedral, Hexagonal, Monoclinic, and Triclinic.
Reason: Gases with the same molar mass diffuse at the same rate. Molar mass of CO₂ = 44 g/mol. Molar mass of C₃H₈ (Propane) = (3 x 12) + (8 x 1) = 44 g/mol.
Reason: For positive rays (protons/ions), the charge-to-mass ratio (e/m) is inversely proportional to the mass of the gas. Since Oxygen has the highest molar mass (16 g/mol) among the given options, it will have the minimum e/m value.
Reason: In the visible spectrum (VIBGYOR), Violet has the highest frequency and the shortest wavelength (approx 400nm), while Red has the longest wavelength (approx 700nm).
Reason: According to the First Law of Thermodynamics, ∆E = q + w. At constant volume, w = 0, so ∆E = qᵥ. Therefore, the change in internal energy is exactly equal to the heat supplied (400 J).
Reason: A high Equilibrium Constant (Kc) indicates that the product concentration is much higher than the reactant concentration at equilibrium, meaning the reaction has proceeded significantly toward completion.
Reason: According to Le Chatelier's Principle, adding heat to an exothermic reaction (which releases heat) shifts the equilibrium to the left (reverse) to absorb the excess heat.
Reason: An open system can exchange both matter (e.g., breathing, eating) and energy (heat loss) with its surroundings. The human body does both.
Reason: Activation Energy (Eₐ) is the difference between Threshold Energy and the average internal energy of reactants. 31 - 12 = 19 kJ/mol.
Reason: Benzene has 6 C-C sigma bonds, 6 C-H sigma bonds (Total 12 σ), and 3 alternating pi bonds in the ring.
Reason: Both have 4 bonding pairs and 0 lone pairs on the central atom (sp³ hybridization), resulting in a tetrahedral geometry with bond angles of 109.5°.
Reason: Dalton's Law of Partial Pressures applies only to non-reacting gas mixtures. HCl and NH₃ react immediately to form solid ammonium chloride (NH₄Cl), so they do not obey the law.
Reason: Use the density formula derived from the ideal gas law:
Using d=PM/RT
Pressure = 1216/760 = 1.6 atm,
Temperature = 50 + 273 = 323 K,
Molar mass of Cl₂ = 71 g/mol
d = 1.6×71/0.0821×323 ≈ 4.28 g/L
Reason: At STP, density is simply Molar Mass divided by Molar Volume.
$$d = \frac{32 \text{ g}}{22.4 \text{ L}} \approx 1.428 \text{ g/L}$$.
Reason: Leading zeros (0.00) are not significant. The "809" are significant, and trailing zeros after a decimal (00) are also significant. Total = 5.
Reason: At 100°C (standard boiling point), the vapor pressure of water equals standard atmospheric pressure. Standard pressure is 760 torr, 1 atm, or 101325 Pascals (Pa).
Reason: This is the definition of the Monoclinic system: all sides unequal, two angles at 90°, and one angle (usually β) not equal to 90°.
Reason: Mercury has extremely strong metallic bonding, resulting in a much higher surface tension than the hydrogen bonding in water or ethanol.
Reason: According to the Arrhenius equation ($$k = Ae^{-E_a/RT}$$), the rate constant is independent of concentration but changes significantly with temperature.
Reason: True solutions have particles < 1nm. Suspensions have particles > 1000nm. Colloids occupy the intermediate range (1nm to 1000nm).
Reason: Metals like copper have a regular, repeating lattice structure, making them crystalline. Glass, plastics, and amorphous silica lack this long-range order.
Reason: 1 mole at STP = 22.4 dm³.
$$1.5 \text{ moles} \times 22.4 \text{ dm}^3/\text{mol} = 33.6 \text{ dm}^3$$.
Reason: An emulsion is a colloid of two or more liquids that are normally immiscible. Mayonnaise is a liquid-in-liquid colloid (oil in water/vinegar) stabilized by egg yolk.
Reason: Using Graham's Law: Rate ratio = √(molar mass of methane / molar mass of helium) = √(16 / 4) = √4 = 2.
Reason: Using the Combined Gas Law: Rate of volume change ∝ (temperature / pressure) ⇒ So, V₂/V₁=(2T/0.5P)=4V
Reason: The unit for a First Order rate constant (or specific rate constant) is s⁻¹, because the concentration terms cancel out in the rate law expression.
Reason: Nitrogen has a triple bond (N≡N). In any triple bond, there is always one sigma (σ) bond and two pi (π) bonds.
Reason: Both CH₄ and CCl₄ have 4 bonding pairs and 0 lone pairs on the central atom, giving them both a Tetrahedral geometry.
Reason: In both Benzene and Ethylene (Ethene), the carbon atoms are sp² hybridized, leading to a planar (trigonal planar around each carbon) geometry.
Reason: Critical temperature is the maximum temperature at which a gas can be liquefied by pressure alone. These parameters are essential for the Liquefaction of gases.
Reason: According to the equation, 1 mole of C produces 1 mole of CO₂. Therefore, 2 moles of C will produce 2 moles of CO₂.
Volume = 2 x 22.4 dm³ = 44.8 dm³.
Reason: Isomorphism is the phenomenon where different substances exist in the same crystalline/geometrical form. Both NaCl and MgO have a cubic crystal structure.
Reason: For an s-orbital, the azimuthal quantum number (l) is 0. Since the magnetic quantum number (mₗ) ranges from -l to +l, for an s-orbital, mₗ can only be 0. Therefore, it cannot have mₗ = -1.
Reason: A very high Equilibrium Constant (Kc) indicates that at equilibrium, the concentration of products is much larger than reactants, meaning the reaction has gone nearly to completion.
Reason: Amphoteric oxides can react with both acids and bases. Zinc oxide (ZnO) and Aluminum oxide (Al₂O₃) are classic examples. CaO is basic, while B₂O₃$ and SiO₂ are acidic.
Reason: In BCl₃, Boron has 3 bond pairs. In C₂H₄ (Ethene) and C₆H₆ (Benzene), each carbon is bonded to 3 other atoms. All these cases require sp² hybridization to form a trigonal planar arrangement.
Reason: H₂S (2 bonds + 2 lone pairs), NH₃ (3 bonds + 1 lone pair), and CH₄ (4 bonds) all have a steric number of 4, requiring sp³ hybridization.
Reason: All these molecules are linear. BeCl₂ has 2 bond pairs, CO₂ has 2 double bonds (2 sigma regions), and C₂H₂ has a triple bond (1 sigma + 2 pi). All utilize sp hybridization.
Reason: Na₂CO₃ is formed from NaOH (Strong Base) and H₂CO₃ (Weak Acid). This results in a basic salt.
(NaNO₃ is Strong-Strong; NH₄Cl is Weak-Strong).
Reason: At STP, the number of moles is directly proportional to volume (n = V / 22.4). Since 1 dm³ is the smallest volume provided, it contains the minimum number of moles (1/22.4 = 0.044 moles.
Reason: A convex meniscus (like Mercury) occurs when the liquid molecules are more attracted to each other (cohesion) than to the glass walls (adhesion). This causes the liquid to "pull away" from the surface.
Reason: Colligative properties depend only on the number of solute particles. Atmospheric pressure is an external environmental factor, not a property of the solution itself.
Reason: Combustion is always an exothermic process, meaning heat is always released (∆H < 0). Formation and reaction can be either endothermic or exothermic.
Reason: NaCl has a face-centered cubic (FCC) structure. Calculation:
Cl⁻ ions: (8 x 1/8 corners) + (6 x 1/2 faces) = 4.
Na⁺ ions: (12 x 1/4 edges) + (1 x 1 center}) = 4.
📝 ✍️ Smart Answers of Section B 🔹 Attempt 9 Short Qs ➡️4 marks each 🎯 36 marks
Q2 (i)
Define any 4 of the following:
Stoichiometry, Molar volume, Mole, Percentage yield, Exponential notation,
Significant figure, Dipole moment, Tyndal effect, Anisotropy, Cleavage plane,
Symmetry, Crystal growth, Transition temperature, Chemical equilibrium,
Gay-Lussac Law, Bond order, Debye, Unit cell, Bond energy, Colloid,
Dispersion forces, Activation energy and Lattice energy, Rate expression,
Rate constant, Critical temperature, Allotropy, Limiting reactant,
Molar heat of vaporization, Dispersion force, Viscosity, Surface tension,
Molar heat of fusion.
Answer
Stoichiometry ⚖️
It is the branch of chemistry that deals with quantitative relationships
between reactants and products in a chemical reaction as per the balanced
equation.
Molar Volume 📦
It is the volume occupied by 1 mole of a gas at STP (273 K, 1 atm) equal to
22.4 L. It is calculated as molar mass divided by gas density.
Mole 🧮
SI unit for the amount of substance, equal to 6.02 × 10²³ particles
(atoms, molecules, ions). A mole is the gram atomic mass or gram molecular
or gram formula mass of any substance which contains 6.02 × 10²³ particles.
Percentage Yield 📊
The ratio of practical yield to theoretical yield is referred as percent yield. It measures efficiency of a reaction,
% Yield = (Actual yield/Theoretical yield) × 100
Exponential / Scientific Notation 🔢
It is the shorthand mathematical expression to write very large or very small numbers using powers of 10. Example: Xⁿ where X is multiplied itself by n times.
Significant Figures ✨
These are digits in a number known with certainty plus one uncertain last digit, showing the reliability of measurement.
Gay-Lussac’s Law of Combining Volumes 🔬
Gases react in simple whole number volume ratios at the same temperature and pressure.
Limiting Reactant ⛔
It is the reactant that is completely consumed first giving least number of moles of product, limiting the amount of product formed.
Dipole Moment (μ) ⚡
It is a measure of molecule polarity showing extent or tendency of a polar molecule to turn or orient in an electric field, calculated as the product of charge × distance between poles.
Debye (D) 💡
The dipole moment of all polar molecules are found in the limit of 10⁻¹⁸ esu.cm (10⁻²⁰ e.s.u-m or 10⁻³⁰ C-m). This quantity is known as debye (D) after the name of introducer (Peter Debye). 1 debye = 1 x 10⁻¹⁸ esu.cm
Bond Order 🔗
It is the number of bonds between two atoms, or half the difference between bonding and antibonding electrons:
Bond order = (bonding electrons – antibonding electrons)/2
Bond Energy 🔥
It is the energy required to break 1 mole of a specific bond in a molecule (kJ/mol). Alternatively, the bond energy can also be defined as the energy released in formation of a bond from its free atoms to form a molecule.
Allotropy 🔄
The polymorphism in elements is called allotropy. It is the existence of an element in two or more crystalline forms in the same state with different physical but same chemical properties.
Anisotropy ↔️
Property of crystals showing different physical properties in different directions.
Crystal Growth 🌱
It is the formation of crystals when a hot saturated solution cools slowly.
Cleavage Plane 🪓
It is plane along which a crystal breaks into smaller crystals of identical shape.
Symmetry 🔄
It is the regular repetition of crystal faces or edges when rotated around an axis.
Transition Temperature 🌡️
It is temperature where two allotropes of the same element coexist in equilibrium.
Unit Cell 🏠
It is the smallest repeating structural unit of a crystal lattice showing 3D pattern possessing a definite geometric shape.
Lattice Energy ⚡
Energy released when 1 mole of gaseous ions form a crystal (negative) or required to break it into ions (positive).
Molar Heat of Fusion (∆Hfusion)🔥➡️💧
It is the heat required to convert 1 mole of solid completely into liquid at melting point without changing temperature.
Molar Heat of Vaporization (∆Hᵥ or ΔHᵥₐₚ)💧➡️💨
It is the heat required to convert 1 mole of liquid into gas at boiling point without changing temperature.
Chemical Equilibrium ⚖️
It is the state of a reversible reaction (in a closed vessel) where the rate of forward reaction equals the rate of reverse reaction with no observable change in concentrations.
Colloid / Colloidal Solution 🥛
It is a solution or mixture with particles 2–1000 nm, intermediate between true solution and suspension (e.g., starch, albumin).
Tyndall Effect 💡
It is the process of the scattering of light by colloidal particles.
Dispersion Forces / London Forces (LDF) 🧲
These are weak short-range temporary dipole-induced dipole attractions, present in all molecules; stronger for larger or elongated molecules. It is the weakest IMF. (Bond energy of LDF = 1-10 kJ/mol or 2 kcal/mol).
Critical Temperature 🌡️
It is the highest temperature at which a gas can be liquefied by pressure.
Rate Constant / Specific Rate Constant ⏱️
It is the proportionality constant in the rate law and it is the ratio between rate of reaction and the product of molar concentration of reactants, giving the reaction rate for unit concentration of reactants.
Activation Energy (Eₐ) ⚡
It is the extra energy (average K.E. + excess energy) required by reacting molecules to reach the threshold for reaction possess in order to form the product:
Eₐ = Threshold energy – Average internal energy
Rate Expression / Rate Law 📝
Experimentally determined relation showing reaction rate as a function of reactant concentrations with each term raised to some power:
Rate = k[A]ᵡ [B]ᵞ
Viscosity 🌀
It is the resistance of a liquid to flow. A more viscous liquid (like honey) flows slower than a less viscous one (like water).
Surface Tension 🌊
Surface tension is the perpendicular force acting per unit length along the surface of a liquid. Surface tension is the surface energy required to increase per unit area of a liquid.
Write down 2 differences between any TWO of the following:
➡️ Solution, colloids and suspension
➡️ Crystalline & amorphous solid
➡️ σ‑bond and π‑bond
➡️ Isomorphism and polymorphism
➡️ Ideal and non‑ideal solution
➡️ Polar and non‑polar bond
➡️ Continuous and line spectrum
➡️ Molecularity & reaction order
➡️ Lyman and Balmer series
➡️ VBT and MOT
➡️ Hydrophobic and Hydrophilic molecules
➡️ Positive and negative catalyst
➡️ BMO and AMO
Answer
| Solution 💧 | Colloid 🥛 | Suspension 🌪️ |
|---|---|---|
| Particle size <1 nm 🔬 | Particle size 1–100 nm 🧫 | Particle size >1000 nm 👀 |
| Homogeneous ✅ | Homogeneous or heterogeneous ⚖️ | Heterogeneous ❌ |
| Clear & transparent 👓 | Cloudy but uniform ☁️ | Cloudy & settles 🏜️ |
| Transparent but often colored🔍🎨 | Translucent & often opaque but can be Transparent 🌫️🔒✨ | Often opaque, but can be transparent⬛👀✨ |
| Cannot be separated ❌ | Hard to separate ⚠️ | Can be separated 🧹 |
| Particles invisible 🔍 | Particles Seen via ultramicroscope 🔬 | Particles Visible to naked eye 👁️ |
| Do not scatter light ✨ | Scatter light (Tyndall effect) 💡 | Scatter light, not transparent 🌫️ |
| Particles Pass through filter 🧻 | Particles Do Not pass filter 🚫 | Particles Pass through filter 🧻 |
| Crystalline Solids 💎 | Amorphous Solids ⚡ |
|---|---|
| True solids ✅ | Pseudo solids / supercooled liquids ⚠️ |
| Regular, ordered structure 🏗️ | Random, irregular structure 🌀 |
| Definite geometric shape 🔷 | Irregular shape ❌ |
| Long-range order 🌐 | Short-range order 🏝️ |
| Anisotropic ↔️ | Isotropic 🔵 |
| Symmetrical ✅ | Unsymmetrical ❌ |
| Sharp melting point 🌡️ | No fixed melting point ⏳ |
| Smooth surfaces when cut ✂️ | Irregular surfaces ✂️ |
| Definite and characteristic heat of fusion ❄️🔥 | Indefinite heat of fusion 🌡️❓ |
| Regular cleavage 🪓 | Irregular cleavage ⚡ |
| When cut with a sharp edged tool, they split into two pieces and new surfaces are plain and smooth | When cut with a sharp edged tool, they split into two pieces with irregular surfaces |
| Examples: Cu, Ag, NaCl, alums, sugar 🏆 | Examples: Glass, rubber, plastics 🧱 |
| Sigma (σ) ➕ | Pi (π) ➖ |
|---|---|
| Formed by head-on overlap 🔄 | Formed by sidewise overlap ↔️ |
| Electron density along bond axis 🧲 | Electron density above & below axis ☁️ |
| Single bond only 1️⃣ | Double/triple bonds 2️⃣/3️⃣ |
| Free rotation 🔄 | No free rotation 🚫 |
| No nodal plane ❌ | Has nodal plane ✔️ |
| Localized bond 📍 | Localized or delocalized 🔄 |
| Strong bond 💪 | Weak bond ⚡ |
| Symmetrical electron cloud ☁️ | Unsymmetrical electron cloud 🌩️ |
| Decides bond direction & length 📏 | Modifies bond, not direction ✏️ |
| Forms first (prior to pi bonds)✅ | Forms after sigma ⏳ |
| Breaks after pi ⏳ | Breaks first ⚡ |
| Both s and p orbitals take part in sigma bond 🧬 | Only p orbitals in pi bond formation 🧬 |
| Can use hybrid orbitals 🌀 | Only unhybridized orbitals ❌ |
| Large overlap 📦 | Small overlap 📦 |
| Decides Shapes molecule 🔷 | Modifies bond, not shape ✏️ |
| Isomorphism 🔷 | Polymorphism ⚡ |
|---|---|
| Similar crystal structure of different compounds 🏗️ | Different crystal forms of same compound 🔄 |
| The crystal shape of isomorphic compounds are identical to each other | The crystal shape of polymorphic substances are different from to each other |
| Crystal shapes identical 🔹 | Crystal shapes different 🔸 |
| Concerns 2+ compounds 2️⃣+ | Concerns 1 compound 1️⃣ |
| Not seen in elements ❌ | Seen in elements ✅ |
| Atomic ratios same ⚖️ | Atomic ratios may vary 🔀 |
| Physical & chemical properties different ⚡ | Physical properties differ, chemical same ⚖️ |
|
Examples: NaF and MgO (Cubic) K₂SO₄ and K₂SeO₄ (Orthorhombic) |
Examples: Sulphur (Rhombic and monoclinic) CaCO₃ (Calcite and Aragonite) |
| Ideal Gas 💨 | Real Gas 🌬️ |
|---|---|
| Particles point-sized, no volume 🔹 | Particles finite volume 📏 |
| Collisions elastic ⚡ | Collisions non-elastic 🐢 |
| No intermolecular forces ❌ | Has intermolecular forces ⚡ |
| Hypothetical gas ❓ | Exists in real environment 🌎 |
| High pressure 📈 | Lower pressure 📉 |
| No interactions ⚪ | Interacts with others 🔗 |
| Obeys ideal gas equation; PV = nRT ✅ | Obeys van der Waal’s eq; (P+(an²/V²)(V−nb)=nRT 🔧 |
| Polar Bond ⚡ | Non-Polar Bond 💨 |
|---|---|
| Formed between different atoms 🧪 | Formed between similar atoms 🧬 |
| ΔEN > 0.5–1.7 ⚡ | ΔEN < 0.5 ⚪ |
| Electrons unevenly shared ⚖️ | Electrons evenly shared ✅ |
| Partial ionic character 🧲 | Pure covalent 🔗 |
| Partial charges (poles) +/– ⚡ | No charges (poles) ❌ |
| May/may not be polar ⚖️ | Non-polar ❌ |
| May have dipole moment 🧲 | Zero dipole moment 0️⃣ |
| Electron cloud distorted 🌩️ | Electron cloud undistorted ⚪ |
| Net dipole present 🧲 | Net dipole absent ❌ |
| Continuous Spectrum 🌈 | Line Spectrum ✨ |
|---|---|
| From white light ☀️ | From excited gas/metal 🔥 |
| Many colours (7) blend 🌈 | Sharp individual lines ✨ |
| No dark spaces ❌ | Dark spaces present ⚫ |
| Polychromatic light 🌈 | Monochromatic light 🎯 |
| Broad wavelength range 🌊 | Limited wavelength 🎯 |
| Cannot determine element structure ❌ | Determines element structure ✅ |
| Incandescent lamp gives it 💡 | Atomic emission gives it 🔥 |
| S. # | Balmer Series 🌈 | Lyman Series 🔥 |
|---|---|---|
| 1. | Visible region 👀 | Ultraviolet region 🌞❌ |
| 2. | Energy lower ⚡ | Energy higher 🔥 |
| 3. | Wavelength 3500–7000 Å 📏 | Wavelength <3500 Å 📐 |
| 4. | Falling orbit nf = 2 ⬇️ | Falling orbit nf = 1 ⬇️ |
| # | Valence Bond Theory (VBT)🧪 | Molecular Orbital Theory (MOT)🌌 |
|---|---|---|
| 1. | Proposed by Heitler & London 👨🔬 | Proposed by F. Hund and R.S. Mulliken 👨🔬 |
| 2. | Orbitals retain identity 🌀 | Orbitals lose identity ❌ |
| 3. | Atomic orbitals monocentric ⚪ | Molecular orbitals polycentric ⚫ |
| 4. | Based on atomic orbitals 🧬 | Based on molecular orbitals 🔗 |
| 5. | Shows hybrid orbitals 🔀 | Shows bonding & antibonding MOs ⚡ |
| 6. | Some valence electrons unshared ⚪ | All valence electrons bonded ✅ |
| 7. | Resonance important 🔄 | Resonance not applicable ❌ |
| 8. | Simpler calculations ✏️ | Complex calculations 📊 |
| 9. | Fails to explain O₂ paramagnetism ❌ | Explains O₂ paramagnetism 🧲 |
| 10. | Fails for H₂⁺ ❌ | Explains H₂⁺ ✅ |
| 11. | Bonds localized 📌 | Electrons delocalized 🌐 |
| 12. | Predicts molecular shape 🔺 | Predicts electron arrangement ⚛️ |
| Reaction Order 🔢 | Molecularity 🔬 |
|---|---|
| Affected by concentration changes 📏 | Total reactants molecules taking part ⚛️ |
| Sum of powers in rate law ✏️ | Sum of coefficients in equation 📐 |
| Experimentally determined 🧪 | Theoretical concept 📘 |
| Can be 0–3 ⚡ | Always 1–3 🔹 |
| Can be zero ✅ | Cannot be zero ❌ |
| Can be fractional ⚖️ | Always whole number 1️⃣,2️⃣,3️⃣ |
| Gives reaction path info 🔍 | Does not give mechanism info ❌ |
| Changes with conditions 🌡️ | Fixed for reaction ✅ |
| Its value cannot exceed than three because in a single collision, the concentration of only 3 molecules alters. | Its value can be more than three but high molecularity reactions are very rare. |
| Natural Radioactivity 🌿 | Artificial Radioactivity ⚡ |
|---|---|
| Spontaneous, heavy nuclei >83 🪨 | Induced, small nuclei ⚡ |
| In it, α, β, γ rays are produced. | In it, sub-atomic particles e.g. neutron, proton, positron are produced. |
| Uncontrollable ❌ | Controllable ✅ |
| Spontaneous ✅ | Induced ❌ |
| Usually heavy elements 🪨 | Can be lighter elements ⚡ |
| U, Th, Ra etc. are the examples of elements that undergo natural radioactivity. | Be, B, Al etc. are the examples of elements that undergo artificial radioactivity. |
| Cannot speed up/slow down ❌ | Controlled by particle speed ⚡ |
| Self-disintegration 🔄 | Induced disintegration ⚡ |
| α, β, γ emissions 🔭 | Emits subatomic particles (n, p, e⁺) ⚛️ |
| Examples: U, Th, Ra 🧪 | Examples: Be, B, Al 🔬 |
| Positive Catalyst ➕ | Negative Catalyst ➖ |
|---|---|
| Increases reaction rate ⚡ | Decreases reaction rate 🐢 |
| Also called accelerator 🚀 | Also called retarder or inhibitor ⏳ |
| Lowers activation energy 🔥 | Increases activation energy 🔥 |
| Widely used to speed up processes | Used to prevent unwanted reactions |
| Provides alternative low-energy pathway | Suppresses reaction pathway |
| Example: MnO₂ in decomposition of H₂O₂ | Example: Glycerol in decomposition of H₂O₂ |
| BMO (Bonding Molecular Orbital) 🧲 | AMO (Antibonding Molecular Orbital) ⚡ |
|---|---|
| Constructive overlap of atomic orbitals | Destructive overlap of atomic orbitals |
| Lower than atomic orbitals ⬇️ | Higher than atomic orbitals ⬆️ |
| Electron density between nuclei 👫 | Electron density outside nuclei 🌌 |
| Stabilizes molecule ✅ | Destabilizes molecule ❌ |
| Promotes bond formation 🔗 | Weakens or Opposes bond ❌🔗 |
| Shorter Bond Length | Longer Bond Length |
| Increases bond order | Decreases bond order |
| Notation: σ, π | Notation: σ*, π* |
Q2 (ii)
State Hund’s rule of multiplicity, Aufbau principle and Pauli exclusion principle. Write the values of four quantum numbers for the valence electrons of He and Mg (Z = 12). Also write down the electronic configuration for ground states of each of the following:
Zn, S²⁻, Cr (24), Cu (29), Fe³⁺ (26), Br⁻ (35), Mo (42), Ag (Z = 47), Pd (Z = 46), Ca²⁺ (Z = 20), Cl⁻, Sr²⁺ (38)
Answer
Quantum Principles ⚛️
✨Pauli’s Exclusion Principle ❌
➡️ An orbital can hold maximum 2 electrons with opposite spins. No two electrons have the same 4 quantum numbers.
✨Aufbau Principle 🏗️
➡️ Electrons fill orbitals progressively in increasing energy order, starting from 1s.
✨Hund’s Rule ✨
➡️ In degenerate orbitals (p, d, f), electrons occupy maximum half-filled orbitals with parallel spins.
Quantum Numbers of Valence Electrons of He (1s² or 1s↿⇂)🔢
1st Electron………. ➡️n = 1 ➡️l (n−1) = 0 ➡️m (2l + 1) = 0 ➡️s or ms = + ½
2nd Electron………➡️ n = 1 ➡️ l (n−1) = 0 ➡️ m (2l + 1) = 0 ➡️ s or ms = – ½
Quantum Numbers of Valence Electrons of Mg (3s² or 3s↿⇂)🔢
1st Electron………. ➡️n = 3 ➡️l (n−1) = 0 ➡️m (2l + 1) = 0 ➡️s or ms = + ½
2nd Electron………➡️ n = 3 ➡️ l (n−1) = 0 ➡️ m (2l + 1) = 0 ➡️ s or ms = – ½
Ground State Electron Configurations of Given Species 🧬
➡️Zn (30) ➡️30ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰ 4s² or [Ar] 3d¹⁰ 4s²
➡️S²⁻(16+2=18)➡️18ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶ or [Ar]
➡️Cr (24) ➡️24ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d⁵ 4s¹ or [Ar] 3d⁵ 4s¹
➡️Cu (29) ➡️29ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰ 4s¹ or [Ar] 3d¹⁰ 4s¹
➡️Fe³⁺(26–3=23)➡️23ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d⁵ or [Ar] 3d⁵
➡️Br⁻ (35+1=36)➡️36ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰ 4s² 4p⁶ or [Kr] 3d¹⁰ 4s² 4p⁶
➡️Mo (42) ➡️42ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰ 4s² 4p⁶, 4d⁵ 5s¹ or [Kr] 4d⁵ 5s¹
➡️Ag (47) ➡️47ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰ 4s² 4p⁶, 4d¹⁰ 5s¹ or [Kr] 4d¹⁰ 5s¹
➡️Pd (46) ➡️46ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰ 4s² 4p⁶, 4d¹⁰ or [Kr] 4d¹⁰
➡️Cl⁻ (17+1=18) ➡️18ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶ or [Ar]
➡️Ca²⁺ (20–2=18) ➡️18ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶ or [Ar]
➡️Sr²⁺ (38–2=36) ➡️36ē —→ 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰ 4s² 4p⁶ or [Kr] 3d¹⁰ 4s² 4p⁶
What is meant by dipole moment? Give its mathematical formulae and different units. On what factors does it depend? Which of the following molecules have dipole moment? In each case, give a reason for your answer: CO₂, CHCl₃, SCl₂, H₂O, CCl₄
Answer
Dipole Moment (μ) ⚡
➡️ Dipole Moment (μ) is the tendency of a polar molecule to orient in an electric field.
➡️ Bond moment: moment of a single bond.
➡️ Molecular or Net dipole moment: vector sum of all bond moments.
➡️ Mathematically it is the product of magnitude of charge at each pole and the inter-nuclear distance (bond distance).
Formula 📐
Dipole moment (μ) = Charge x distance between two opposite charge OR μ=e⋅d
Units 🧪
📌 SI: Coulomb-meter (C·m)
📌 CGS: esu·cm (electrostatic unit-centimeter)
📌 Common: Debye (D) (One debye corresponds to the dipole moment produced when a charge equal to the electronic charge is separated from an equal and opposite charge by a distance of about 10⁻¹⁰ m
📌 Relation: 1D = 3.34 ×10⁻³⁰ C⋅m (1D ≈ 10⁻¹⁸e.s.u-cm or 10⁻²⁰e.s.u-m)
Factors Affecting Dipole Moment ⚖️
❶ Polarity of molecule
📌 Greater electronegativity difference (ΔEN) → Higher polarity → Larger μ
📌 ΔEN ∝ Polarity ∝ μ
❷ Molecular Geometry
📌 Dipole moment depends on bond polarity + shape (for polyatomic molecule).
📌 Vector sum of bond moments determines overall μ.
Predicing Dipole Moment of Given Molecules with Reason 🧩
Dipole Moments of Given Molecules
Dipole Moment of CO₂➡️❌ Zero➡️ Linear geometry → bond moments cancel (vector sum = 0)
Dipole Moment of CHCl₃➡️✅ Some➡️ Unsymmetrical tetrahedral → bond moments do not cancel
Dipole Moment of SCl₂➡️✅ Some➡️ Angular shape → vector sum ≠ 0
Dipole Moment of H₂O➡️✅ Some➡️ Angular shape → vector sum ≠ 0
Dipole Moment of CCl₄➡️❌ Zero➡️ Symmetrical tetrahedral → bond moments cancel(vector sum = 0)
Q2 (iii)
What are quantum numbers and orbitals? Give a brief account of 4 quantum numbers. Write all possible values of l, m and s for n = 2 and n = 3. Draw the shape of different orbitals with l = 2. Arrange the following orbitals according to Wiswesser rule: 4f, 3d, 4s, 6p, 7s, 5d.
Answer
✨Quantum Numbers (QNs) 🔢
➡️ Constant integral numbers that describe electron energy, orbital shape, orientation, and spin.
➡️ Derived from Schrödinger’s Wave Equation.
✨Orbital 🌀
➡️A region around nucleus where an electron is most likely to be found.
➡️Has fixed energy and spatial distribution.
✨4 Quantum Numbers 🧩
🔥Principal QN (n)➡️ Energy level / shell ➡️ Values = 1,2,3…
🔥Azimuthal QN (ℓ)➡️ Subshell / shape➡️ Values = 0 → s, 1 → p, 2 → d, 3 → f
🔥Magnetic QN (m or mₗ) ➡️ Orientation in space➡️ Values = -ℓ → + ℓ
🔥Spin QN (s or mₛ)➡️ Spin of electron ➡️ Values = +½, -½
✨All Possible Values of ℓ, m and s for n=2 and n=3
Quantum Numbers Table
| Energy Level (n) | Sub-energy Level (ℓ) | Orientation of Orbitals (m) | Spin (s) |
|---|---|---|---|
| 2 (L-shell) | 0 (2s) | 0 (s) | +½, -½ |
| 2 (L-shell) | 1 (2p) | -1, 0, +1 (pₓ pᵧ p_z) | +½, -½ |
| 3 (M-shell) | 0 (3s) | 0 (s) | +½, -½ |
| 3 (M-shell) | 1 (3p) | -1, 0, +1 (pₓ pᵧ p_z) | +½, -½ |
| 3 (M-shell) | 2 (3d) | -2, -1, 0, +1, +2 (dₓᵧ, dₓz, dᵧz, dₓ²₋ᵧ², d_z²) | +½, -½ |